Asked by Joshua
I'm having a lot of trouble with this problem:
Sketch the graph and show all local extrema and inflections.
f(x)= (x^(1/3)) ((x^2)-175)
I graphed the function on my graphing calculator and found the shape.
I also found the first derivative:
(7/3)(x^(4/3)) - (175/3)(x(-2/3))
But the number I found for x (when I set the deriv. to 0 to find the critical points) was 1.71, but this doesn't correspond to a min or a max. Then I plugged the equation of the first derivative in onto my calculator, and the corresponding y value of 1.71 was NOT zero.... furthermore, it seemed like there was a horizontal asymptote at 0, because the derivative never was 0.
I couldn't find an inflection point, because the second derivative never equaled zero.
Does this seem right-- that there is no local max, min, OR inflection point? If this is wrong, can you help me find the right values? I sketched the graph on my paper but I still haven't found the min, max, or inflection points.
Sketch the graph and show all local extrema and inflections.
f(x)= (x^(1/3)) ((x^2)-175)
I graphed the function on my graphing calculator and found the shape.
I also found the first derivative:
(7/3)(x^(4/3)) - (175/3)(x(-2/3))
But the number I found for x (when I set the deriv. to 0 to find the critical points) was 1.71, but this doesn't correspond to a min or a max. Then I plugged the equation of the first derivative in onto my calculator, and the corresponding y value of 1.71 was NOT zero.... furthermore, it seemed like there was a horizontal asymptote at 0, because the derivative never was 0.
I couldn't find an inflection point, because the second derivative never equaled zero.
Does this seem right-- that there is no local max, min, OR inflection point? If this is wrong, can you help me find the right values? I sketched the graph on my paper but I still haven't found the min, max, or inflection points.
Answers
Answered by
Reiny
I had the same first derivative as you did
y' = (7/3)(x^(4/3)) - (175/3)(x(-2/3))
= (1/3)x^(-2/3)[7x^2 - 175] by factoring
setting this equal to zero, ...
the first factor of (1/3)x^(-2/3) yields no solution but
7x^2-175=0 gives me
x = ± 5
my second derivative was (28/9)x^(1/3) + (350/9)x^(-5/3) which when set to zero has no answer.
so according to the math, there should be 2 points of max/min, but no point of inflection.
Does that fit in with your sketch?
I did not sketch the graph.
y' = (7/3)(x^(4/3)) - (175/3)(x(-2/3))
= (1/3)x^(-2/3)[7x^2 - 175] by factoring
setting this equal to zero, ...
the first factor of (1/3)x^(-2/3) yields no solution but
7x^2-175=0 gives me
x = ± 5
my second derivative was (28/9)x^(1/3) + (350/9)x^(-5/3) which when set to zero has no answer.
so according to the math, there should be 2 points of max/min, but no point of inflection.
Does that fit in with your sketch?
I did not sketch the graph.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.